a) $n_{Ba} = \dfrac{41,1}{137} = 0,3(mol)$
$n_{H_2SO_4} = \dfrac{200.4,9\%}{98} = 0,1(mol)$
$Ba + H_2SO_4 \to BaSO_4 + H_2(1)$
$Ba + 2H_2O \to Ba(OH)_2 + H_2(2)$
Theo PTHH : $n_{H_2} = n_{Ba} = 0,3(mol)$
$V_{H_2} = 0,3.22,4 = 6,72(lít)$
b)
Theo PTHH :
$n_{BaSO_4} = n_{H_2SO_4} = 0,1(mol)$
$n_{Ba(OH)_2} = n_{Ba} - n_{BaSO_4} = 0,3 - 0,1 = 0,2(mol)$
$m_{dd\ sau\ pư} = m_{Ba} + m_{dd\ H_2SO_4} - m_{H_2} - m_{BaSO_4}$
$= 41,1 + 200 - 0,3.2 - 0,1.233 = 217,2(gam)$
$C\%_{Ba(OH)_2} = \dfrac{0,2.171}{217,2}.100\% = 15,74\%$